Math / Calculus

Question18. Surface Area All edges of a cube are expanding at rate of 6 centimeters per second. How fast is the surface area changing when each edge is (a) 2 centimeters and (b) 10 centimeters?

Studdy Solution

STEP 1

1. The cube has edges of length $s$ .
2. The rate of change of the edge length $\frac{ds}{dt} = 6$ cm/s.
3. The surface area $A$ of a cube is given by $A = 6s^2$ .
4. We need to find the rate of change of the surface area $\frac{dA}{dt}$ .

STEP 2

1. Express the surface area of the cube as a function of the edge length.
2. Differentiate the surface area with respect to time.
3. Calculate the rate of change of the surface area for given edge lengths.

STEP 3

The surface area $A$ of a cube is given by the formula: $A = 6s^2$

STEP 4

Differentiate $A = 6s^2$ with respect to time $t$ to find $\frac{dA}{dt}$ : $\frac{dA}{dt} = \frac{d}{dt}(6s^2) = 12s \frac{ds}{dt}$

STEP 5

Substitute $\frac{ds}{dt} = 6$ cm/s into the differentiated equation: $\frac{dA}{dt} = 12s \times 6 = 72s$

STEP 6

(a) When $s = 2$ cm, substitute $s = 2$ into $\frac{dA}{dt} = 72s$ : $\frac{dA}{dt} = 72 \times 2 = 144 \text{ cm}^2/\text{s}$

STEP 7

(b) When $s = 10$ cm, substitute $s = 10$ into $\frac{dA}{dt} = 72s$ : $\frac{dA}{dt} = 72 \times 10 = 720 \text{ cm}^2/\text{s}$
The rate of change of the surface area is: (a) $144 \text{ cm}^2/\text{s}$ when $s = 2$ cm. (b) $720 \text{ cm}^2/\text{s}$ when $s = 10$ cm.

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Question18. Surface Area All edges of a cube are expanding at rate of 6 centimeters per second. How fast is the surface area changing when each edge is (a) 2 centimeters and (b) 10 centimeters?

Studdy Solution

STEP 1

1. The cube has edges of length $s$ .
2. The rate of change of the edge length $\frac{ds}{dt} = 6$ cm/s.
3. The surface area $A$ of a cube is given by $A = 6s^2$ .
4. We need to find the rate of change of the surface area $\frac{dA}{dt}$ .

STEP 2

1. Express the surface area of the cube as a function of the edge length.
2. Differentiate the surface area with respect to time.
3. Calculate the rate of change of the surface area for given edge lengths.

STEP 3

The surface area $A$ of a cube is given by the formula: $A = 6s^2$

STEP 4

Differentiate $A = 6s^2$ with respect to time $t$ to find $\frac{dA}{dt}$ : $\frac{dA}{dt} = \frac{d}{dt}(6s^2) = 12s \frac{ds}{dt}$

STEP 5

Substitute $\frac{ds}{dt} = 6$ cm/s into the differentiated equation: $\frac{dA}{dt} = 12s \times 6 = 72s$

STEP 6

(a) When $s = 2$ cm, substitute $s = 2$ into $\frac{dA}{dt} = 72s$ : $\frac{dA}{dt} = 72 \times 2 = 144 \text{ cm}^2/\text{s}$

STEP 7

(b) When $s = 10$ cm, substitute $s = 10$ into $\frac{dA}{dt} = 72s$ : $\frac{dA}{dt} = 72 \times 10 = 720 \text{ cm}^2/\text{s}$
The rate of change of the surface area is: (a) $144 \text{ cm}^2/\text{s}$ when $s = 2$ cm. (b) $720 \text{ cm}^2/\text{s}$ when $s = 10$ cm.

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Question18. Surface Area All edges of a cube are expanding at rate of 6 centimeters per second. How fast is the surface area changing when each edge is (a) 2 centimeters and (b) 10 centimeters?

Studdy Solution

STEP 1

1. The cube has edges of length s s s.2. The rate of change of the edge length dsdt=6 \frac{ds}{dt} = 6 dtds​=6 cm/s.3. The surface area A A A of a cube is given by A=6s2 A = 6s^2 A=6s2.4. We need to find the rate of change of the surface area dAdt \frac{dA}{dt} dtdA​.

STEP 2

1. Express the surface area of the cube as a function of the edge length.2. Differentiate the surface area with respect to time.3. Calculate the rate of change of the surface area for given edge lengths.

STEP 3

The surface area A A A of a cube is given by the formula: A=6s2 A = 6s^2 A=6s2

STEP 4

Differentiate A=6s2 A = 6s^2 A=6s2 with respect to time t t t to find dAdt \frac{dA}{dt} dtdA​: dAdt=ddt(6s2)=12sdsdt \frac{dA}{dt} = \frac{d}{dt}(6s^2) = 12s \frac{ds}{dt} dtdA​=dtd​(6s2)=12sdtds​

STEP 5

Substitute dsdt=6 \frac{ds}{dt} = 6 dtds​=6 cm/s into the differentiated equation: dAdt=12s×6=72s \frac{dA}{dt} = 12s \times 6 = 72s dtdA​=12s×6=72s

STEP 6

(a) When s=2 s = 2 s=2 cm, substitute s=2 s = 2 s=2 into dAdt=72s \frac{dA}{dt} = 72s dtdA​=72s: dAdt=72×2=144 cm2/s \frac{dA}{dt} = 72 \times 2 = 144 \text{ cm}^2/\text{s} dtdA​=72×2=144 cm2/s

STEP 7

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1. The cube has edges of length $s$ .
2. The rate of change of the edge length $\frac{ds}{dt} = 6$ cm/s.
3. The surface area $A$ of a cube is given by $A = 6s^2$ .
4. We need to find the rate of change of the surface area $\frac{dA}{dt}$ .

1. Express the surface area of the cube as a function of the edge length.
2. Differentiate the surface area with respect to time.
3. Calculate the rate of change of the surface area for given edge lengths.

The surface area $A$ of a cube is given by the formula: $A = 6s^2$

Differentiate $A = 6s^2$ with respect to time $t$ to find $\frac{dA}{dt}$ : $\frac{dA}{dt} = \frac{d}{dt}(6s^2) = 12s \frac{ds}{dt}$

Substitute $\frac{ds}{dt} = 6$ cm/s into the differentiated equation: $\frac{dA}{dt} = 12s \times 6 = 72s$

(a) When $s = 2$ cm, substitute $s = 2$ into $\frac{dA}{dt} = 72s$ : $\frac{dA}{dt} = 72 \times 2 = 144 \text{ cm}^2/\text{s}$